How Can We Solve for 'a' and 'b' in This Simultaneous Equation?

[anemone]

Solve the simultaneous equation and find the values of $a$ and $b$.

$\dfrac{a+b}{a+\sqrt{1+a^2}}+b^2=0$

$\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3$

 

[Ackbach]

Spoiler

By inspection, $a=0, b=-1$ solves the system. There may be other solutions.

 

[anemone]

Thanks for participating, Ackbach!:)

Ahem... I can only give you at most half credit because your answer isn't complete!(Wondering):p

My solution:

Spoiler

We're given

$\dfrac{a+b}{a+\sqrt{1+a^2}}+b^2=0$	$\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3$
If we let $f(a)=a+\sqrt{1+a^2}$, we see that	Observe that
1. $f(a) \ne 0$, 2. $f(a) > 0$ for all $a$ since $\;\;\;\;$when $a \rightarrow \ -\infty$, $f(a) \rightarrow 0$ $\;\;\;\;$when $a \rightarrow \infty$, $f(a) \rightarrow \infty$ View attachment 1678	$\sqrt{1+a^2} \ge 1$ for all $a$ and hence we have $2\sqrt{1+a^2} \ge 2$. Thus, $\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3$ is true when the following inequality holds. $\dfrac{a^2}{b^2}+b^2 \le 1$ $a^2+b^4 \le b^2$ $b^4 \le b^2-a^2$ $b^4 \le (a+b)(b-a)$
Thus, $\dfrac{a+b}{a+\sqrt{1+a^2}}+b^2=0$ is feasible iff $a+b<0$---(*).	Since $b^4 \ge 0$, and $a+b<0$(from *), we know $b-a<0$---(**).

These two inequalities (*) and (**) suggest that $b<0$.

Now, if we let

$k=a+\sqrt{1+a^2}=f(a)>0$, $\rightarrow a=\dfrac{k^2-1}{2k}$

and substitute these two into the equation $\dfrac{a+b}{a+\sqrt{1+a^2}}+b^2=0$, we get

$\dfrac{\dfrac{k^2-1}{2k}+b}{k}+b^2=0$

$\dfrac{\dfrac{k^2-1+2bk}{2k}}{k}+b^2=0$

$\dfrac{k^2-1+2bk}{2k^2}+b^2=0$

$k^2-1+2bk+2k^2b^2=0$

$(1+2b^2)k^2-(2b)k-1=0$

Solving it for $k$ by the quadratic formula, we obtain

$k=\dfrac{b\pm \sqrt{3b^2+1}}{1+2b^2}$

Since $k>0$, and $b<0$, we can say at this point that $k=\dfrac{b+ \sqrt{3b^2+1}}{1+2b^2}$, which in turn tells us we have only one value of $x$ as the answer.

Now, the only answer to this problem, like Ackbach has mentioned is $(x,y)=(0, -1)$.

 

[Albert1]

Solve the simultaneous equation and find the values of $a$ and $b$.

$\dfrac{a+b}{a+\sqrt{1+a^2}}+b^2=0$

$\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3$

$\dfrac{a+b}{a+\sqrt{1+a^2}}+b^2=0---(1)$

$\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3---(2)$
from (2) $b\neq 0$
from (1) it is easy to see a+b<0
from (2)$\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3>\dfrac{a^2}{b^2}+2\sqrt{a^2}+b^2>3\times \left |a \right |\sqrt[3]{2}$(AP>GP)
$\therefore 1>\left | a \right |\sqrt[3]{2}$
if $a\in Z$ then a=0 and from (1)b=-1

 

[anemone]

$\dfrac{a+b}{a+\sqrt{1+a^2}}+b^2=0---(1)$

$\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3---(2)$
from (2) $b\neq 0$
from (1) it is easy to see a+b<0
from (2)$\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3>\dfrac{a^2}{b^2}+2\sqrt{a^2}+b^2>3\times (\pm) a\sqrt[3]{2}$(AP>GP)
$\therefore 1>(\pm a)\sqrt[3]{2}$
if $a\in Z$ then a=0 and from (1)b=-1

Hi Albert,

Thanks for showing me another method to solve this problem and thanks too for participating. But I don't understand why there is a plus minus sign in front of the cube root of 2, after one applied the AM-GM to the terms $\dfrac{a^2}{b^2}, 2\sqrt{a^2}, b^2$:

$\dfrac{a^2}{b^2}+2\sqrt{a^2}+b^2>3(\pm \sqrt[3]{2})$

I see that even if $1>\pm\sqrt[3]{2}(a)$ is true, the solution set for $a$ that I can get from here is $-\dfrac{1}{\sqrt[3]{2}}<a<\dfrac{1}{\sqrt[3]{2}}$, but not what you have concluded... I'm sure I am missing something here, and I hope you will enlighten me.

 

[Albert1]

$\dfrac{a+b}{a+\sqrt{1+a^2}}+b^2=0---(1)$

$\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3---(2)$
from (2) $b\neq 0$
from (1) it is easy to see a+b<0
from (2)$\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3>\dfrac{a^2}{b^2}+2\sqrt{a^2}+b^2>3\times \left |a \right |\sqrt[3]{2}$(AP>GP)
$\therefore 1>\left | a \right |\sqrt[3]{2}$
we get a=0 and from (1)b=-1

we get a=0 (if a is an integer),and from (1) b=-1
if a is not an integer ----(please wait )

 

[Opalg]

[sp]This is a follow-up to anemone's solution above. She shows that $b<0$, and also that $a^2+b^4 \leqslant b^2$, from which $a^2 \leqslant b^2(1-b^2).$ It follows that $|a| < |b|$ and also that $|b| \leqslant 1.$ Therefore $-1\leqslant b<0.$

Multiplying out the fractions, we can write the original equations as $$1+b + ab^2 + b^2c = 0, \qquad(1)$$ $$a^2 + 2b^2c + b^4 - 3b^2 = 0, \qquad (2)$$ where $c = \sqrt{1+a^2}$. Multiply (1) by $2$ and subtract (2) (to eliminate $c$): $$2a + 2b + 2ab^2 - a^2 - b^4 + 3b^2 = 0.$$ Write that as a quadratic in $a$ and solve it: $$a^2 - 2(1+b^2)a + (b^4 - 3b^2 - 2b) = 0,\qquad a = 1 + b^2 - \sqrt{5b^2 + 2b + 1} \qquad (3)$$ (we have to take the negative square root in (3) because $|a| < |b|$).

Using the definition of $c$, write (2) as $(c^2 - 1) + 2b^2c + b^4 - 3b^2 = 0$. Write that as a quadratic in $c$ and solve it: $$c^2 + 2cb^2 + (b^4 - 3b^2 - 1) = 0, \qquad c = -b^2 + \sqrt{3b^2+1} \qquad (4)$$ (this time, we have to take the positive square root because $c\geqslant 1$).

Now substitute (4) into (1) and solve for $a$: $$a+b + ab^2 + b^2\bigl(-b^2 + \sqrt{3b^2+1}\bigr), \qquad a = \frac{b^4 - b^2\sqrt{3b^2+1} - b}{1+b^2}. \qquad(5)$$

The next step is to equate the two expressions for $a$ ((3) and (5)), which gives an equation for $b$. After a small amount of simplification, this becomes: $$f(b) \overset{\text{def}}{=} 1+ b + 2b^2 + b^2\sqrt{3b^2+1} - (1+b^2) \sqrt{5b^2 + 2b + 1} = 0.$$
I don't see any hope of solving that equation, but a plot of the graph of $f(b)$ shows that its only zero in the interval $-1\leqslant b<0$ is when $b=-1$ (click on View attachment 1690 to see the printout).[/sp]

 

[anemone]

[sp]This is a follow-up to anemone's solution above. She shows that $b<0$, and also that $a^2+b^4 \leqslant b^2$, from which $a^2 \leqslant b^2(1-b^2).$ It follows that $|a| < |b|$ and also that $|b| \leqslant 1.$ Therefore $-1\leqslant b<0.$

Multiplying out the fractions, we can write the original equations as $$1+b + ab^2 + b^2c = 0, \qquad(1)$$ $$a^2 + 2b^2c + b^4 - 3b^2 = 0, \qquad (2)$$ where $c = \sqrt{1+a^2}$. Multiply (1) by $2$ and subtract (2) (to eliminate $c$): $$2a + 2b + 2ab^2 - a^2 - b^4 + 3b^2 = 0.$$ Write that as a quadratic in $a$ and solve it: $$a^2 - 2(1+b^2)a + (b^4 - 3b^2 - 2b) = 0,\qquad a = 1 + b^2 - \sqrt{5b^2 + 2b + 1} \qquad (3)$$ (we have to take the negative square root in (3) because $|a| < |b|$).

Using the definition of $c$, write (2) as $(c^2 - 1) + 2b^2c + b^4 - 3b^2 = 0$. Write that as a quadratic in $c$ and solve it: $$c^2 + 2cb^2 + (b^4 - 3b^2 - 1) = 0, \qquad c = -b^2 + \sqrt{3b^2+1} \qquad (4)$$ (this time, we have to take the positive square root because $c\geqslant 1$).

Now substitute (4) into (1) and solve for $a$: $$a+b + ab^2 + b^2\bigl(-b^2 + \sqrt{3b^2+1}\bigr), \qquad a = \frac{b^4 - b^2\sqrt{3b^2+1} - b}{1+b^2}. \qquad(5)$$

The next step is to equate the two expressions for $a$ ((3) and (5)), which gives an equation for $b$. After a small amount of simplification, this becomes: $$f(b) \overset{\text{def}}{=} 1+ b + 2b^2 + b^2\sqrt{3b^2+1} - (1+b^2) \sqrt{5b^2 + 2b + 1} = 0.$$
I don't see any hope of solving that equation, but a plot of the graph of $f(b)$ shows that its only zero in the interval $-1\leqslant b<0$ is when $b=-1$ (click on View attachment 1690 to see the printout).[/sp]

Thank you Opalg for participating and showing us another way to tackle the problem.

After reading your post, I realized I have provided a false solution. My apologies.

I will stand in the corner with my nose against the wall for making the silly assumption as follows: (Angry)

In $k=\dfrac{b\pm \sqrt{3b^2+1}}{1+2b^2}$, since $k>0$, and $b<0$, we can say at this point that $k=\dfrac{b+ \sqrt{3b^2+1}}{1+2b^2}$, which in turn tells us we have only one value of $x$ as the answer".

 

[Albert1]

Solve the simultaneous equation and find the values of $a$ and $b$.

$\dfrac{a+b}{a+\sqrt{1+a^2}}+b^2=0$---(1)

$\dfrac{a^2}{b^2}+2\sqrt{1+a^2}+b^2=3---(2)$

from (1) a+b<0
from (2) $a^2<b^2,\therefore b<a,\,\, (a-b)>0$$\therefore b<0$
the range of a : -1<a<1---(3)
the range of b:$-1\leq b<0$---(4)
let $\sqrt {1+a^2}=1+x(x\geq 0)$
we get $a^2=x^2+2x$
from (2) $\dfrac {x^2+2x}{b^2}+2+2x+b^2=3$
$\dfrac {x^2+2x+2b^2x+b^4}{b^2}=1=\dfrac{b^2}{b^2}$
$(x^2+2x)+b^2(b^2+2x)=b^2$
compare both side we have:
$(x^2+2x)=0 ,\,\, and (b^2+2x)=1$
x=0 or x=-2 (deleted)
so we get a=0 and b=-1 or (b=1 also deleted)
$\therefore$ (a,b)=(0,-1) is the only solution

 

[kaliprasad]

from (1) a+b<0
from (2) $a^2<b^2,\therefore b<a,\,\, (a-b)>0$$\therefore b<0$
the range of a : -1<a<1---(3)
the range of b:$-1\leq b<0$---(4)
let $\sqrt {1+a^2}=1+x(x\geq 0)$
we get $a^2=x^2+2x$
from (2) $\dfrac {x^2+2x}{b^2}+2+2x+b^2=3$
$\dfrac {x^2+2x+2b^2x+b^4}{b^2}=1=\dfrac{b^2}{b^2}$
$(x^2+2x)+b^2(b^2+2x)=b^2$
compare both side we have:
$(x^2+2x)=0 ,\,\, and (b^2+2x)=1$
x=0 or x=-2 (deleted)
so we get a=0 and b=-1 or (b=1 also deleted)
$\therefore$ (a,b)=(0,-1) is the only solution

$(x^2+2x)+b^2(b^2+2x)=b^2$
compare both side we have:
$(x^2+2x)=0 ,\,\, and (b^2+2x)=1$

this statement is not correct. This is correct if it is true for any b and not a particular b.

 

[Albert1]

$(x^2+2x)+b^2(b^2+2x)=b^2$
compare both side we have:
$(x^2+2x)=0 ,\,\, and (b^2+2x)=1$

this statement is not correct. This is correct if it is true for any b and not a particular b.

please note the range of a,and b
in fact b=$\pm 1,\,\, and\,\, b=\pm \sqrt 5$
but b=1, and b=$\pm \sqrt 5$ must be deleted

 

[Opalg]

from (1) a+b<0
from (2) $a^2<b^2,\therefore b<a,\,\, (a-b)>0$$\therefore b<0$
the range of a : -1<a<1---(3)
the range of b:$-1\leq b<0$---(4)
let $\sqrt {1+a^2}=1+x(x\geq 0)$
we get $a^2=x^2+2x$
from (2) $\dfrac {x^2+2x}{b^2}+2+2x+b^2=3$
$\dfrac {x^2+2x+2b^2x+b^4}{b^2}=1=\dfrac{b^2}{b^2}$
$(x^2+2x)+b^2(b^2+2x)=b^2$
compare both side we have:
$(x^2+2x)=0 ,\,\, and (b^2+2x)=1$
x=0 or x=-2 (deleted)
so we get a=0 and b=-1 or (b=1 also deleted)
$\therefore$ (a,b)=(0,-1) is the only solution

Very nice solution. I would prefer to simplify the final part of it by writing $x^2+2x+2b^2x+b^4 = b^2$ as $(x+b^2)^2 + 2x = b^2$. Since $x\geqslant0$, the left side is clearly greater than the right unless $x=0$.

That proof relies entirely on information from the second of the two given equations; the first one is not needed at all!

 

[kaliprasad]

Very nice solution. I would prefer to simplify the final part of it by writing $x^2+2x+2b^2x+b^4 = b^2$ as $(x+b^2)^2 + 2x = b^2$. Since $x\geqslant0$, the left side is clearly greater than the right unless $x=0$.

That proof relies entirely on information from the second of the two given equations; the first one is not needed at all!

some how I am missing the clearly greater than the right part.

if b^2 < 1 then b^4 < b^2 in that case x could be positive. I can neither prove it nor disprove it.

any help in this regards

 

[Opalg]

some how I am missing the clearly greater than the right part.

if b^2 < 1 then b^4 < b^2 in that case x could be positive. I can neither prove it nor disprove it.

any help in this regards

You are quite right – I was confusing $b^2$ with $b^4$ at some point, and my "simplification" does not work. (Doh)